What is the set of real values of x that satisfies the quadratic inequality 2x^2 - 5x - 3 … — JAMB UTME

What is the set of real values of $x$ that satisfies the quadratic inequality $2x^2 - 5x - 3 \le 0$ ?

-\frac{1}{2} \le x \le 3

Factoring

2x^2 - 5x - 3

yields

(2x + 1)(x - 3) \le 0

. Setting the factors to zero gives roots

x = -1/2

and

x = 3

. Because the coefficient of

x^2

is positive, the quadratic curve opens upwards and is less than or equal to zero in the closed interval between the roots, resulting in

-\frac{1}{2} \le x \le 3

Factor the quadratic expression

2x^2 - 5x - 3

(2x + 1)(x - 3) \le 0

Finding the factors helps identify the critical points (roots) of the inequality.

Determine the critical points by setting each factor equal to zero

x = -\frac{1}{2}

and

x = 3

The critical points divide the real number line into test intervals.

Determine the region where

(2x + 1)(x - 3) \le 0

-\frac{1}{2} \le x \le 3

Since the quadratic coefficient is positive (

2 > 0

), the parabola opens upwards, so the function values are less than or equal to zero between the two roots.

Solving Quadratic Inequalities by Factorisation

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